$ F = \left[\begin{array}{rr}-1 & 3 \\ -1 & 5 \\ 5 & 3\end{array}\right]$ $ E = \left[\begin{array}{rr}-2 & 4 \\ 5 & 4\end{array}\right]$ What is $ F E$ ?
Answer: Because $ F$ has dimensions $(3\times2)$ and $ E$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ F E = \left[\begin{array}{rr}{-1} & {3} \\ {-1} & {5} \\ \color{gray}{5} & \color{gray}{3}\end{array}\right] \left[\begin{array}{rr}{-2} & \color{#DF0030}{4} \\ {5} & \color{#DF0030}{4}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ F$ , with the corresponding elements in column $j$ of the second matrix, $ E$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ F$ with the first element in ${\text{column }1}$ of $ E$ , then multiply the second element in ${\text{row }1}$ of $ F$ with the second element in ${\text{column }1}$ of $ E$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-1}\cdot{-2}+{3}\cdot{5} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ F$ with the corresponding elements in ${\text{column }1}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{-2}+{3}\cdot{5} & ? \\ {-1}\cdot{-2}+{5}\cdot{5} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ F$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{-1}\cdot{-2}+{3}\cdot{5} & {-1}\cdot\color{#DF0030}{4}+{3}\cdot\color{#DF0030}{4} \\ {-1}\cdot{-2}+{5}\cdot{5} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-1}\cdot{-2}+{3}\cdot{5} & {-1}\cdot\color{#DF0030}{4}+{3}\cdot\color{#DF0030}{4} \\ {-1}\cdot{-2}+{5}\cdot{5} & {-1}\cdot\color{#DF0030}{4}+{5}\cdot\color{#DF0030}{4} \\ \color{gray}{5}\cdot{-2}+\color{gray}{3}\cdot{5} & \color{gray}{5}\cdot\color{#DF0030}{4}+\color{gray}{3}\cdot\color{#DF0030}{4}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}17 & 8 \\ 27 & 16 \\ 5 & 32\end{array}\right] $